Mole+Unit+Resources

Substances react according to the **Law of Definite Proportions**. The following balanced chemical equation shows that two atoms of aluminum react with three molecules of iodine to form two formula units of aluminum iodide.
 * Measuring Matter**

2Al(s) + 3I2(s) --> 2AlI3(s)

If you wanted to carry out this reaction, how could you measure the correct numbers of particles of aluminum and iodine? Fortunately, you can count particles by measuring mass. Suppose you have a sample of an element, and the mass of the sample in grams is numerically equal to the atomic mass of the element. Scientists have discovered that this mass of an element contains 6.02 x 10^23 atoms (molecules, formula units, or ions) of that element. This number is called **Avogadro's number**. Avogadro's number of particles is called a **mole** of particles. The mole is the SI base unit used to measure the amount of a substance and is defined as the number of particles in exactly 12 grams of pure carbon-12. For the purpose of dealing with moles, the simplest unit of any substance is called a representative particle. The particle may be an atom, a molecule, a formula unit, or an ion.

How many molecules are in 2.25 moles of bromine?
 * Example Problem: //Converting Moles to Number of Particles//**

Bromine is an element that consists of diatomic molecules. Therefore, one mole of bromine consists of 6.02 x 10^23 Br2 molecules. To find the number of Br2 molecules present in 2.25 mol, multiply the number of moles by Avogadro's number.

number of Br2 molecules =

2.25 mol x __ 6.02 x 10^23 molecules __ 1 mol

number of Br2 molecules = 1.35 x 10^24 molecules

Calculate the number of moles in a sample of sodium bromide (NaBr) that contains 2.88 x 10^23 formula units.
 * Example Problem: //Converting Number of Particles to//** //Moles//

Because 1 mol NaBr = 6.02 x 10^23 formula units NaBr, you can see that 2.88 x 10^23 formula units is less than one mole of NaBr.

moles of NaBr =

2.88 x 10^23 formula units x __1 mole NaBr__ 6.02 x 10^23 formula units

moles of NaBr = 0.478 mol NaBr

**Mass and the Mole** One mole of an element consists of 6.02 x 10^23 atoms of that element. The mass of a mole of any substance is called the **molar mass** of the substance. For example, the molar mass of an element is numerically equal to the atomic mass of the element, but expressed in grams.

molar mass of a substance = __grams of the substance__ 1 mole of the substance This relationship can be used to convert between mass and moles.

A roll of copper wire has a mass of 848 g. How many moles of copper are in the roll? Use the atomic mass of copper given on the periodic table to apply a conversion factor to the mass given in the problem.
 * Example Problem: //Converting Mass to Moles//**

moles of Cu = grams Cu x __1 mole Cu__ grams Cu

moles of Cu = 848 g Cu x __1 mole Cu__ = 13.3 mol Cu 64 g Cu

Note that the conversion factor is arranged so that g Cu cancel, leaving only mol Cu, the desired quantity.

Calculate the mass of 0.625 moles of calcium.
 * Example Problem: //Converting Moles to//** **//Mass//**

Use the molar mass of calcium to apply a conversion factor to the number of moles given in the problem. According to the periodic table, the atomic mass of calcium is 40 amu, so the molar mass of calcium is 40 g.

mass of Ca = mol Ca x __40 g Ca__ = 25 g Ca 1 mol Ca

The conversion factor is arranged so that mol Ca cancel, leaving g Ca, the desired quantity.

Calculate the number of atoms in 4.77 g lead.
 * Example Problem: //Converting Mass to Number of Particles//**

To find the number of atoms in the sample, you must first determine how many moles are in 4.77 g lead. According to data from the periodic table, the molar mass of lead is 207.2 g/mol. Apply a conversion factor to convert mass to moles.

moles of Pb = grams Pb x __1 mol Pb__ grams Pb

moles of Pb = 4.77 g Pb x __ 6.02 x 10^23 atom Pb __ 1 mol Pb

atoms of Pb = 0.0230 mol Pb x __ 6.02 x 10^23 atom Pb __ = 1.38 x 10^22 atom Pb 1 mol Pb

You can also convert from number of particles to mass by reversing the procedure above and dividing the number of particles by Avogadro's number to determine the number of moles present.

Recall that a mole is Avogadro's number ( 6.02 x 10^23) of particles of a substance. If the substance is a molecular compound, such as ammonia (NH3), a mole is 6.02 x 10^23 molecules of ammonia. If the substance is an ionic compound, such as baking soda (sodium hydrogen carbonate, NaHCO3), a mole is 6.02 x 10^23 formula units of sodium hydrogen carbonate. In either case, a mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound. For example, a mole of ammonia (NH3) consists of one mole of nitrogen atoms and three moles of hydrogen atoms.
 * Moles of Compounds**

**Molar Mass of a Compound** The molar mass of a compound is the mass of a mole of the representative particles of the compound. Because each representative particle is composed of two or more atoms, the molar mass of the compound is found by adding the molar masses of all of the atoms in the representative particle. In the case of NH3, the molar mass equals the mass of one mole of nitrogen atoms plus the mass of three moles of hydrogen atoms.

molar mass of NH3 = molar mass of N + 3(molar mass of H) molar mass of NH3 = 14g + 3(1g) = 17 g/mol

You can use the molar mass of a compound to convert between mass and moles, just as you used the molar mass of elements to make these conversions.

**Example Problem: //Converting Mass of a Compound to Moles//** At 4.0 degrees C, water has a density of 1 g/mL. How many moles of water are in 1 kg of water (1 L at 4 degrees C)?

Before you can calculate moles, you must determine the molar mass of water (H2O). A mole of water consists of two moles of hydrogen atoms and one mole of oxygen atoms.

molar mass H2O = 2(molar mass H) + molar mass O molar mass H2O = 2(1 g) + 16 g = 18 g/mol

Now you can use the molar mass of water as a conversion factor to determine moles of water. Notice that 1 kg is converted to 1 x 10^3 g for the calculation.

moles of H2O = grams H2O x __1mole H2O__ 18 g H2O

moles of H2O = 1 x 10^3 g H2O x __1 mole H2O__ 18 g H2O 0.05551 x 10^3 mol H2O moles of H2O = 55.51 mol H2O

Notice that the result must have four significant figures because the mass and volume data in the problem were given to four significant figures.

Recall that every chemical compound has a definite composition - a composition that is always the same wherever that compound is found. The composition of a compound is usually stated as the percent by mass of each element in the compound.
 * Empirical and Molecular Formulas**

The percent of an element in a compound can be found in the following way.
 * Percent Composition**

% by mass of element = __mass of element in 1 mol compound__ x 100 molar mass of compound

The example problem below shows you how to determine the **percent composition** of a compound, which is the percent by mass of each element in the compound.

Determine the percent composition of calcium chloride (CaCl2).
 * Example Problem: //Calculating Percent// //Composition//**

First, analyze the information available from the formula. A mole of calcium chloride consists of one mole of calcium ions and two moles of chloride ions.

Next, gather molar mass information from the atomic masses on the periodic table. To the mass of one mole of CaCl2, a mole of calcium ions contributes 40 g, and two moles of chloride ions contribute 2 x 35 g = 70 g for a total molar mass of 110 g/mol for CaCl2.

Finally, use the data to suet up a calculation to determine the percent by mass of each element in the compound. The percent by mass of calcium and chlorine in CaCl2 can be calculated as follows.

% Ca in CaCl2 = __40 g Ca__ x 100 = 36.112% Ca 110 g CaCl2

% Cl in CaCl2 = __70 g Cl__ x 100 = 63.888% Cl 110 g CaCl2

As a check, be sure that the percentages add up to 100%. In this case, the percentages add up to 100%.

You can use percent composition data to help identify an unknown compound by determining its empirical formula. The **empirical formula** is the simplest whole-number ratio of atoms of elements in the compound. In many cases, the empirical formula is the actual formula for the compound. For example, the simplest ratio of atoms of sodium to atoms of chlorine in sodium chloride is 1 atom Na : 1 atom Cl. So, the empirical formula of sodium chloride Na1Cl1, or NaCl, which is the true formula for the compound. The following example problem will show you how to determine empirical formulas.
 * Empirical Formulas**

The percent composition of an unknown compound is found to be 38.43% Mn, 16.80% C, and 44.77% O. Determine the compound's empirical formula.
 * Example Problem: //Empirical Formula from Percent Composition// **

Because //percent// means "parts per hundred parts," assume that you have 100 g of the compound. Then calculate the number of moles of each element in the 100 g of compound. The number of moles of manganese may be calculated as follows.

moles of Mn (in 100 g) = grams Mn (in 100 g) x __1 mol Mn__ molar mass Mn

moles of Mn (in 100 g) = 38.43 g Mn x __1 mol Mn__ = 0.6995 mol Mn 54.938 g Mn

By following the same pattern, the number of moles of carbon and oxygen per 100 g sample may be calculated.

moles of C (in 100 g) = 16.80 g C x __1 mol C__ = 1.399 mol C 12.011 g C moles of O (in 100 g) = 44.77 g O x __1 mol O__ = 2.798 mol O 15.99 g O The results show the following relationship. mol Mn: mol C: mol O = 0.6995 : 1.339 : 2.798

To obtain the simplest whole-number ratio of moles, divide each number of moles by the smallest number of moles.

moles of Mn = __0.6995 mol Mn__ = 1 mol Mn 0.6995 moles of C = __1.399 mol C__ = 2 mol C 0.6995 moles of O = __2.798 mol O__ = 4 mol O 0.6995 The empirical formula for the compound is MnC2O4

For many compounds, the empirical formula is not the true formula. For example, the empirical formula for acetic acid is CH2O. Chemists have learned, though, that acetic acid is a molecule with the formula C2H4O2, which is the molecular formula for acetic acid. A **molecular formula** tells the exact number of atoms of each element in a molecule or formula unit of a compound. Notice that the molecular formula for acetic acid, (C2H4O2), has exactly twice as many atoms of each element as the empirical formula (CH2O). The molecular formula for a compound is always a whole-number multiple of the empirical formula.
 * Molecular Formulas**

In order to determine the molecular formula for an unknown compound, you must know the molar mass of the compound in addition to its empirical formula. Then you can compare the molar mass of the compound with the molar mass represented by the empirical formula as shown in the following example problem.

Maleic acid is a compound that is widely used in the plastics and textiles industries. The composition of maleic acid is 41.39% carbon, 3.47% hydrogen, and 55.14% oxygen. Its molar mass is 116.1 g/mol. Calculate the molecular formula for maleic acid.
 * Example Problem: //Determining a Molecular Formula//**

Start by determining the emrical formula for a compound.

moles of C (in 100 g) = 41.39 g C x __1 mol C__  = 3.446 mol C 12.011 g C moles of H (in 100 g) = 3.47 g H x  __1 mol H__  = 3.442 mol H 1.008 g H moles of O (in 100 g) = 55.14 g O x __ 1 mole O __ = 3.446 mol O 15.999 g O The numbers of moles of C, H, and O are nearly equal, so it is not necessary to divide through by the smallest value. You can see by inspection that the smallest whole-number ratio is 1C : 1H : 1O, and the empirical formula is CHO.

Next, calculate the molar mass represented by the formula CHO. Here, the molar mass is the sum of the masses of one mole of each element. molar mass CHO = 12.011 g + 1.008 g + 15.999 g molar mass CHO = 29.018 g/mol

As stated in the problem, the molar mas of maleic acid is known to be 116.1 g/mol. To determine the molecular formula for maleic acid, calculate the whole number multiple, //n//, to apply to its empirical formula.

//n// = __116.1 g/mol maleic acid__ = 4.001 29.018 g/mol CHO

This calculation shows that the molar mass of maleic acid is four times the molar mass of its empirical formula CHO. Therefore, the molecular formula must have four times as many atoms of each element as the empirical formula. Thus, the molecular formula is (CHO)4 = C4H4O4.